[stenzer]

Hi guys,

I calculated the cooling of a plate by a turbulent forced convection analytically. Now I want to check my results with COMSOL 4.3 but I have some troubles, maybe someone would be so kind and help me.

My plate is 0.075 m long, to keep the plate at +4 °C i need a power of ~130 W. The cooling fluid is air and has a temperature of -30 °C with a "inflow" - velocity of 45 m/s.

I tried to model my plate in COMSOL with a rectangle ( 0.075 m x 0.005 m ), this plate is in the upper right corner of an "larger" rectangle. The inlet is on the left. The material of my plate is copper.

My first problem is how to fit the power, because for my analytic approach i have just a surface (~0.017 m^2) but no volume, which is necessary for a heat source with a unit of W/m^3. When I try to run the simulation with "Total Power" equal my required 130 W, the result is far away from my analytical one.

I'm thankful for any advice, I'm pretty new at COMSOL and an electrical engineer so I'm not really familiar with themal stuff ;-).

BR

[FMDenaro]

If I understand correctly your question, you should work by setting the heat flux in W/m^2:

q = - k grad T

so that, by fixing q, you set a non-homogeneous Neumann BC.s

dT/dn|wall = -q/k

[stenzer]

Hi,

I don't know if I understand you correctly.

I calculated the power which is needed to keep my plate at a certain temperature (+4°C). So in my opinion I have to model the plate as a heat source and not as heat flux ?!?!????

BR

[FMDenaro]

Originally Posted by stenzer

Hi,

I don't know if I understand you correctly.

I calculated the power which is needed to keep my plate at a certain temperature (+4°C). So in my opinion I have to model the plate as a heat source and not as heat flux ?!?!????

BR

Could you explain your problem by means of a sketch?

[stenzer]

Hi,

here is the link for a sketch, I hope you can read everything.

https://www.dropbox.com/s/ggs1e7wrfvk2y5c/Cooling.jpg

thx

[FMDenaro]

it is a 2D model? therefore, your power for m^3 must be considered for 1m of unit lenght in the third direction (outward the sketch plane) in such a way you have a volume.
I think you could model a mixed fluid-copper problem, so that the heat exchange at wall is due to the heat flux. In the copper region you can solve a standard diffusion equation for the temperature which is governed by a parabolic equation like

dT/dt = k Lap T

[stenzer]

Hi,

I did it like you suggested. But it doesn't fit with my "analytical" result. I already checked my calculations, but in my opinoin they a are right.

At +4°C the kinematic viscosity (nu) of airis v =13.85e-6 m^2/s, Pr = 0.7116 and the heat conductivity k = 24.6e-3 W/mK.

Thus my Reynolds number is: Re = 243682

Because I want to investigate a plate with a plunt edge I assumed a complete turbulent flow, thus I can calculate my Nusselt number by

Nu = [0.037 * Re^(0.8) * Pr]/[1 + 12.7 * sqrt{ 0.037 * Re^-0.2} * (Pr^(2/3) - 1)] = 626.795

Now I can determine my heat transfere coefficient h and the needed power P to keep the surface at +4°C.

h = 205.5 W/(m^2 K)
P = 0.176 m^2 * h * (4°C + 30°C) = 133 W


I think I made no mistake, but maybe someone can find a mistake or give me any advise how to model that in COMSOL. I'm useing Conjugate Heat Transfer and k - epsilon Turbulent Model.

thx

[FMDenaro]

Originally Posted by stenzer

Hi,

I did it like you suggested. But it doesn't fit with my "analytical" result. I already checked my calculations, but in my opinoin they a are right.

At +4°C the kinematic viscosity (nu) of airis v =13.85e-6 m^2/s, Pr = 0.7116 and the heat conductivity k = 24.6e-3 W/mK.

Thus my Reynolds number is: Re = 243682

Because I want to investigate a plate with a plunt edge I assumed a complete turbulent flow, thus I can calculate my Nusselt number by

Nu = [0.037 * Re^(0.8) * Pr]/[1 + 12.7 * sqrt{ 0.037 * Re^-0.2} * (Pr^(2/3) - 1)] = 626.795

Now I can determine my heat transfere coefficient h and the needed power P to keep the surface at +4°C.

h = 205.5 W/(m^2 K)
P = 0.176 m^2 * h * (4°C + 30°C) = 133 W


I think I made no mistake, but maybe someone can find a mistake or give me any advise how to model that in COMSOL. I'm useing Conjugate Heat Transfer and k - epsilon Turbulent Model.

thx

well, standard results of a flow over a flat plate established that a fully turbulent flow should be achieved for Re number greater than yours ... I think you are still in a transitional region and I am doubtful a k-eps model can manage succesfull such a condition

[stenzer]

Hi,

thank you for your fast reply. I already thougt about that, because the turbulent flow can be assumed for Reynolds numbers between 5*10^5 and 10^7. To be honest I thought that would be modeled by setting the boundary/wall condition as Wall function .

I will calculate the model for a laminar flow and check if the result is correct. But my problem is still that I have to know how a plate with a plunt edge would act. Is there a possibility to get results for such a model???

[SergeAS]

The use of the wall function (as shown on your sketch) for k-eps turbulence model assumes that y+ on the plate is in the range of 30-200, but for heat transfer problems the recommended value of y+ ~ 1 and no wall function (for reasonable results)

[stenzer]

Hi SergeAS,

I use conjugate heat transfer, so I think that should be considered by the model automatically!?!?!???

My idea is to calculate and simulate the whole model for a longer plate, thus I get a larger Reeynolds number. And because of the Similitude the result for my smaller plate should be correct. Can I do it that way???

BR

[SergeAS]

Originally Posted by stenzer

Hi SergeAS,

I use conjugate heat transfer, so I think that should be considered by the model automatically!?!?!???

I'm not familiar with this product, but I do not think that the inclusion of the option "conjugate heat transfer" is automatically refine your mesh to y + ~ 1

PS: Take a look http://www.comsol.com/blogs/which-tu...d-application/

Low Reynolds Number k-epsilon
The Low Reynolds number k-epsilon is similar to the k-epsilon model but does not use wall functions; it solves the flow everywhere. It is a logical extension to k-epsilon and shares many of its advantages, but uses more memory. It is often advisable to use the k-epsilon model to first compute a good initial condition for solving the Low Reynolds number k-epsilon model. Since it does not use wall functions, lift and drag forces and heat flux can be modeled with higher accuracy.

[stenzer]

Thank you for the link, I will try it with Low Reynolds number k-epsilon.

thx